Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{z + 3}{-z^2 + 9z} \div \dfrac{-9z - 54}{z^2 - 3z - 54} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{z + 3}{-z^2 + 9z} \times \dfrac{z^2 - 3z - 54}{-9z - 54} $ First factor the quadratic. $a = \dfrac{z + 3}{-z^2 + 9z} \times \dfrac{(z - 9)(z + 6)}{-9z - 54} $ Then factor out any other terms. $a = \dfrac{z + 3}{-z(z - 9)} \times \dfrac{(z - 9)(z + 6)}{-9(z + 6)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (z + 3) \times (z - 9)(z + 6) } { -z(z - 9) \times -9(z + 6) } $ $a = \dfrac{ (z + 3)(z - 9)(z + 6)}{ 9z(z - 9)(z + 6)} $ Notice that $(z + 6)$ and $(z - 9)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ (z + 3)\cancel{(z - 9)}(z + 6)}{ 9z\cancel{(z - 9)}(z + 6)} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $a = \dfrac{ (z + 3)\cancel{(z - 9)}\cancel{(z + 6)}}{ 9z\cancel{(z - 9)}\cancel{(z + 6)}} $ We are dividing by $z + 6$ , so $z + 6 \neq 0$ Therefore, $z \neq -6$ $a = \dfrac{z + 3}{9z} ; \space z \neq 9 ; \space z \neq -6 $